En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Développements limitésFonctions d'une variable réelle/Exercices/Développements limités », n'a pu être restituée correctement ci-dessus.
Calculer le développement limité de
arsinh
(
1
+
2
x
+
3
x
2
)
{\displaystyle \operatorname {arsinh} \left(1+2x+3x^{2}\right)}
Solution
Commençons par calculer le dl en 0 à l'ordre 3 de
arsinh
(
1
+
u
)
{\displaystyle \operatorname {arsinh} (1+u)}
Trigonométrie hyperbolique/Fonctions hyperboliques réciproques ) :
arsinh
′
(
1
+
u
)
=
1
1
+
(
1
+
u
)
2
=
1
2
1
+
u
+
u
2
/
2
=
1
2
(
1
−
1
2
(
u
+
u
2
/
2
)
+
3
8
u
2
)
+
o
(
u
2
)
=
1
2
(
1
−
u
2
+
u
2
8
)
+
o
(
u
2
)
{\displaystyle \operatorname {arsinh} '(1+u)={\frac {1}{\sqrt {1+(1+u)^{2}}}}={\frac {1}{{\sqrt {2}}{\sqrt {1+u+u^{2}/2}}}}={\frac {1}{\sqrt {2}}}\left(1-{\frac {1}{2}}\left(u+u^{2}/2\right)+{\frac {3}{8}}u^{2}\right)+o\left(u^{2}\right)={\frac {1}{\sqrt {2}}}\left(1-{\frac {u}{2}}+{\frac {u^{2}}{8}}\right)+o\left(u^{2}\right)}
arsinh
(
1
+
u
)
=
arsinh
(
1
)
+
1
2
(
u
−
u
2
4
+
u
3
24
)
+
o
(
u
3
)
{\displaystyle \operatorname {arsinh} (1+u)=\operatorname {arsinh} (1)+{\frac {1}{\sqrt {2}}}\left(u-{\frac {u^{2}}{4}}+{\frac {u^{3}}{24}}\right)+o\left(u^{3}\right)}
arsinh
(
1
+
2
x
+
3
x
2
)
=
arsinh
(
1
)
+
1
2
(
2
x
+
3
x
2
−
4
x
2
+
12
x
3
4
+
8
x
3
24
)
+
o
(
x
3
)
=
ln
(
1
+
2
)
+
1
2
(
2
x
+
2
x
2
−
8
x
3
3
)
+
o
(
x
3
)
=
ln
(
1
+
2
)
+
2
x
+
2
x
2
−
4
2
x
3
3
+
o
(
x
3
)
.
{\displaystyle {\begin{aligned}\operatorname {arsinh} \left(1+2x+3x^{2}\right)&=\operatorname {arsinh} (1)+{\frac {1}{\sqrt {2}}}\left(2x+3x^{2}-{\frac {4x^{2}+12x^{3}}{4}}+{\frac {8x^{3}}{24}}\right)+o\left(x^{3}\right)\\&=\ln \left(1+{\sqrt {2}}\right)+{\frac {1}{\sqrt {2}}}\left(2x+2x^{2}-{\frac {8x^{3}}{3}}\right)+o\left(x^{3}\right)\\&=\ln \left(1+{\sqrt {2}}\right)+{\sqrt {2}}x+{\sqrt {2}}x^{2}-{\frac {4{\sqrt {2}}x^{3}}{3}}+o\left(x^{3}\right).\end{aligned}}}
Calculer le développement limité de
(
sin
2
x
−
sin
(
x
2
)
)
(
e
tan
x
−
e
tanh
x
)
{\displaystyle \left(\sin ^{2}x-\sin(x^{2})\right)\left(\operatorname {e} ^{\tan x}-\operatorname {e} ^{\tanh x}\right)}
Solution
Le premier facteur est équivalent à
−
x
4
/
3
{\displaystyle -x^{4}/3}
2
x
3
/
3
{\displaystyle 2x^{3}/3}
9
−
3
=
6
{\displaystyle 9-3=6}
9
−
4
=
5
{\displaystyle 9-4=5}
sin
2
x
−
sin
(
x
2
)
=
(
x
−
x
3
6
+
x
5
120
)
2
−
(
x
2
−
x
6
6
)
+
o
(
x
6
)
=
−
x
4
3
+
19
x
6
90
+
o
(
x
6
)
{\displaystyle \sin ^{2}x-\sin(x^{2})=\left(x-{\frac {x^{3}}{6}}+{\frac {x^{5}}{120}}\right)^{2}-\left(x^{2}-{\frac {x^{6}}{6}}\right)+o\left(x^{6}\right)=-{\frac {x^{4}}{3}}+{\frac {19x^{6}}{90}}+o\left(x^{6}\right)}
e
tan
x
−
e
tanh
x
=
2
x
3
3
+
(
x
+
x
3
3
)
2
−
(
x
−
x
3
3
)
2
2
+
(
x
+
x
3
3
)
3
−
(
x
−
x
3
3
)
3
6
+
o
(
x
5
)
=
2
x
3
+
2
x
4
+
x
5
3
+
o
(
x
5
)
.
{\displaystyle {\begin{aligned}\operatorname {e} ^{\tan x}-\operatorname {e} ^{\tanh x}&={\frac {2x^{3}}{3}}+{\frac {\left(x+{\frac {x^{3}}{3}}\right)^{2}-\left(x-{\frac {x^{3}}{3}}\right)^{2}}{2}}+{\frac {\left(x+{\frac {x^{3}}{3}}\right)^{3}-\left(x-{\frac {x^{3}}{3}}\right)^{3}}{6}}+o\left(x^{5}\right)\\&={\frac {2x^{3}+2x^{4}+x^{5}}{3}}+o\left(x^{5}\right).\end{aligned}}}
(
sin
2
x
−
sin
(
x
2
)
)
(
e
tan
x
−
e
tanh
x
)
=
x
7
9
(
−
1
+
19
x
2
30
)
(
2
+
2
x
+
x
2
)
+
o
(
x
9
)
=
x
7
9
(
−
2
−
2
x
+
4
x
2
15
)
+
o
(
x
9
)
=
−
2
x
7
9
−
2
x
8
9
+
4
x
9
135
+
o
(
x
9
)
.
{\displaystyle {\begin{aligned}\left(\sin ^{2}x-\sin(x^{2})\right)(\operatorname {e} ^{\tan x}-\operatorname {e} ^{\tanh x})&={\frac {x^{7}}{9}}\left(-1+{\frac {19x^{2}}{30}}\right)\left(2+2x+x^{2}\right)+o\left(x^{9}\right)\\&={\frac {x^{7}}{9}}\left(-2-2x+{\frac {4x^{2}}{15}}\right)+o\left(x^{9}\right)\\&=-{\frac {2x^{7}}{9}}-{\frac {2x^{8}}{9}}+{\frac {4x^{9}}{135}}+o\left(x^{9}\right).\end{aligned}}}
Calculer le développement limité de
1
x
2
−
1
(
arcsin
x
)
2
{\displaystyle {\frac {1}{x^{2}}}-{\frac {1}{(\arcsin x)^{2}}}}
Solution
Puisque
1
x
2
−
1
(
arcsin
x
)
2
=
1
−
(
arcsin
x
x
)
−
2
x
2
{\displaystyle {\frac {1}{x^{2}}}-{\frac {1}{(\arcsin x)^{2}}}={\frac {1-\left({\frac {\arcsin x}{x}}\right)^{-2}}{x^{2}}}}
arcsin
x
x
{\displaystyle {\frac {\arcsin x}{x}}}
arcsin
x
{\displaystyle \arcsin x}
arcsin
′
x
=
1
1
−
x
2
=
1
+
x
2
2
+
3
x
4
8
+
5
x
6
16
+
o
(
x
7
)
{\displaystyle \arcsin 'x={\frac {1}{\sqrt {1-x^{2}}}}=1+{\frac {x^{2}}{2}}+{\frac {3x^{4}}{8}}+{\frac {5x^{6}}{16}}+o\left(x^{7}\right)}
arcsin
x
=
x
+
x
3
6
+
3
x
5
40
+
5
x
7
112
+
o
(
x
8
)
{\displaystyle \arcsin x=x+{\frac {x^{3}}{6}}+{\frac {3x^{5}}{40}}+{\frac {5x^{7}}{112}}+o\left(x^{8}\right)}
1
x
2
−
1
(
arcsin
x
)
2
=
1
−
(
1
+
x
2
6
+
3
x
4
40
+
5
x
6
112
+
o
(
x
7
)
)
−
2
x
2
=
2
(
x
2
6
+
3
x
4
40
+
5
x
6
112
)
−
3
(
x
2
6
+
3
x
4
40
)
2
+
4
(
x
2
6
)
3
x
2
+
o
(
x
5
)
=
1
3
+
x
2
15
+
31
x
4
945
+
o
(
x
5
)
.
{\displaystyle {\begin{aligned}{\frac {1}{x^{2}}}-{\frac {1}{(\arcsin x)^{2}}}&={\frac {1-\left(1+{\frac {x^{2}}{6}}+{\frac {3x^{4}}{40}}+{\frac {5x^{6}}{112}}+o\left(x^{7}\right)\right)^{-2}}{x^{2}}}\\&={\frac {2\left({\frac {x^{2}}{6}}+{\frac {3x^{4}}{40}}+{\frac {5x^{6}}{112}}\right)-3\left({\frac {x^{2}}{6}}+{\frac {3x^{4}}{40}}\right)^{2}+4\left({\frac {x^{2}}{6}}\right)^{3}}{x^{2}}}+o\left(x^{5}\right)\\&={\frac {1}{3}}+{\frac {x^{2}}{15}}+{\frac {31x^{4}}{945}}+o\left(x^{5}\right).\end{aligned}}}
Calculer le développement limité de
1
x
−
a
ln
(
1
+
x
)
−
b
e
x
−
1
{\displaystyle {\frac {1}{x}}-{\frac {a}{\ln(1+x)}}-{\frac {b}{\operatorname {e} ^{x}-1}}}
Solution
1
−
a
−
b
x
+
b
−
a
2
−
b
−
a
12
x
−
a
24
x
2
+
o
(
x
2
)
{\displaystyle {\frac {1-a-b}{x}}+{\frac {b-a}{2}}-{\frac {b-a}{12}}x-{\frac {a}{24}}x^{2}+o(x^{2})}
Soit
f
{\displaystyle f}
f
(
x
)
=
a
x
+
b
x
2
+
o
(
x
2
)
{\displaystyle f(x)=ax+bx^{2}+o(x^{2})}
a
≠
0
{\displaystyle a\neq 0}
Elle admet donc, au voisinage de 0, une fonction réciproque possédant un d.l. à l'ordre 2 en 0 :
f
−
1
(
y
)
=
y
a
+
c
y
2
+
o
(
y
2
)
{\displaystyle f^{-1}(y)={\frac {y}{a}}+cy^{2}+o(y^{2})}
Déterminer
c
{\displaystyle c}
Solution
(Pour l'existence, admise ici, du d.l. de la réciproque locale, ainsi qu'un exemple numérique plus poussé, voir par exemple ce « devoir corrigé de MPSI »
x
=
f
−
1
(
f
(
x
)
)
=
a
x
+
b
x
2
a
+
c
(
a
x
)
2
+
o
(
x
2
)
{\displaystyle x=f^{-1}(f(x))={\frac {ax+bx^{2}}{a}}+c(ax)^{2}+o(x^{2})}
c
=
−
b
a
3
{\displaystyle c=-{\frac {b}{a^{3}}}}
Soient
f
(
x
)
=
x
−
x
3
3
!
+
x
5
5
!
+
o
(
x
5
)
{\displaystyle f(x)=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+o(x^{5})}
g
(
x
)
=
x
+
a
x
3
+
b
x
5
+
o
(
x
5
)
{\displaystyle g(x)=x+ax^{3}+bx^{5}+o(x^{5})}
Trouver
a
{\displaystyle a}
b
{\displaystyle b}
f
∘
g
(
x
)
=
x
+
x
5
+
o
(
x
5
)
{\displaystyle f\circ g(x)=x+x^{5}+o(x^{5})}
g
∘
f
{\displaystyle g\circ f}
Soit
f
(
x
)
=
{
cos
x
si
x
≥
0
cosh
−
x
si
x
<
0.
{\displaystyle f(x)={\begin{cases}\cos {\sqrt {x}}&{\text{si }}x\geq 0\\\cosh {\sqrt {-x}}&{\text{si }}x<0.\end{cases}}}
Démontrer que
f
{\displaystyle f}
Solution
Pour
x
≥
0
{\displaystyle x\geq 0}
f
(
x
)
=
∑
k
=
0
n
(
−
1
)
k
x
2
k
(
2
k
)
!
+
o
(
x
2
n
+
2
)
=
∑
k
=
0
n
(
−
x
)
k
(
2
k
)
!
+
o
(
x
n
+
1
)
{\displaystyle f(x)=\sum _{k=0}^{n}{\frac {(-1)^{k}{\sqrt {x}}^{2k}}{(2k)!}}+o({\sqrt {x}}^{2n+2})=\sum _{k=0}^{n}{\frac {(-x)^{k}}{(2k)!}}+o(x^{n+1})}
pour
x
<
0
{\displaystyle x<0}
f
(
x
)
=
∑
k
=
0
n
−
x
2
k
(
2
k
)
!
+
o
(
−
x
2
n
+
2
)
=
{\displaystyle f(x)=\sum _{k=0}^{n}{\frac {{\sqrt {-x}}^{2k}}{(2k)!}}+o({\sqrt {-x}}^{2n+2})=}
Redémontrer le théorème de Taylor-Young par application itérée de la première règle de l'Hôpital.
Donner les d.l. à l'ordre 3 :
en 0 de
cos
×
sin
{\displaystyle \cos \times \sin }
en 0 de
sin
∘
k
:=
sin
∘
sin
∘
⋯
∘
sin
{\displaystyle \sin ^{\circ k}:=\sin \circ \sin \circ \dots \circ \sin }
k
{\displaystyle k}
k -ième itérée
sin
{\displaystyle \sin }
k -ième puissance,
sin
k
x
:=
(
sin
x
)
k
=
o
(
x
3
)
{\displaystyle \sin ^{k}x:=(\sin x)^{k}=o(x^{3})}
k
>
3
{\displaystyle k>3}
en 0 de
(
1
+
x
)
α
{\displaystyle \left(1+x\right)^{\alpha }}
α
{\displaystyle \alpha }
en 1 de
e
x
{\displaystyle \operatorname {e} ^{\sqrt {x}}}
Solution
cos
x
sin
x
=
(
1
−
x
2
2
+
o
(
x
2
)
)
(
x
−
x
3
6
+
o
(
x
3
)
)
{\displaystyle \cos x\sin x=\left(1-{\frac {x^{2}}{2}}+o(x^{2})\right)\left(x-{\frac {x^{3}}{6}}+o(x^{3})\right)}
cos
x
sin
x
=
sin
2
x
2
=
2
x
−
(
2
x
)
3
6
+
o
(
x
3
)
2
{\displaystyle \cos x\sin x={\frac {\sin 2x}{2}}={\frac {2x-{\frac {(2x)^{3}}{6}}+o(x^{3})}{2}}}
cos
x
sin
x
=
x
−
2
x
3
3
+
o
(
x
3
)
{\displaystyle \cos x\sin x=x-{\frac {2x^{3}}{3}}+o(x^{3})}
∀
k
∈
N
∗
sin
∘
k
(
x
)
=
x
−
k
6
x
3
+
o
(
x
3
)
{\displaystyle \forall k\in \mathbb {N} ^{*}\quad \sin ^{\circ k}(x)=x-{\frac {k}{6}}x^{3}+o(x^{3})}
k
=
0
{\displaystyle k=0}
sin
∘
0
=
i
d
R
{\displaystyle \sin ^{\circ 0}=\mathrm {id} _{\mathbb {R} }}
Posons
f
(
x
)
=
(
1
+
x
)
α
{\displaystyle f(x)=\left(1+x\right)^{\alpha }}
f
(
0
)
=
1
{\displaystyle f(0)=1}
f
′
(
x
)
=
α
(
1
+
x
)
α
−
1
{\displaystyle f'(x)=\alpha \left(1+x\right)^{\alpha -1}}
f
″
(
x
)
=
α
(
α
−
1
)
(
1
+
x
)
α
−
2
{\displaystyle f''(x)=\alpha \left(\alpha -1\right)\left(1+x\right)^{\alpha -2}}
f
‴
(
x
)
=
α
(
α
−
1
)
(
α
−
2
)
(
1
+
x
)
α
−
3
{\displaystyle f'''(x)=\alpha \left(\alpha -1\right)\left(\alpha -2\right)\left(1+x\right)^{\alpha -3}}
f
′
(
0
)
=
α
{\displaystyle f'(0)=\alpha }
f
″
(
0
)
=
α
(
α
−
1
)
{\displaystyle f''(0)=\alpha \left(\alpha -1\right)}
f
‴
(
0
)
=
α
(
α
−
1
)
(
α
−
2
)
{\displaystyle f'''(0)=\alpha \left(\alpha -1\right)\left(\alpha -2\right)}
f
(
x
)
=
f
(
0
)
+
f
′
(
0
)
x
+
f
″
(
0
)
2
!
x
2
+
f
‴
(
0
)
3
!
x
3
+
o
(
x
3
)
=
1
+
α
x
+
α
(
α
−
1
)
2
x
2
+
α
(
α
−
1
)
(
α
−
2
)
6
x
3
+
o
(
x
3
)
{\displaystyle f(x)=f(0)+f'(0)x+{\frac {f''(0)}{2!}}x^{2}+{\frac {f'''(0)}{3!}}x^{3}+o\left(x^{3}\right)=1+\alpha x+{\frac {\alpha \left(\alpha -1\right)}{2}}x^{2}+{\frac {\alpha \left(\alpha -1\right)\left(\alpha -2\right)}{6}}x^{3}+o\left(x^{3}\right)}
n : cf. développements limités des fonctions usuelles en zéro ).
Le cas particulier
α
=
1
2
{\displaystyle \alpha ={\frac {1}{2}}}
1
+
h
=
1
+
h
2
−
h
2
8
+
h
3
16
+
o
(
h
3
)
=
1
+
u
{\displaystyle {\sqrt {1+h}}=1+{\frac {h}{2}}-{\frac {h^{2}}{8}}+{\frac {h^{3}}{16}}+o\left(h^{3}\right)=1+u}
u
:=
h
2
−
h
2
8
+
h
3
16
+
o
(
h
3
)
→
0
{\displaystyle u:={\frac {h}{2}}-{\frac {h^{2}}{8}}+{\frac {h^{3}}{16}}+o\left(h^{3}\right)\to 0}
h
→
0
{\displaystyle h\to 0}
en 0 de la fonction exponentielle à l'ordre 3 (à nouveau, voir développements limités des fonctions usuelles en zéro ) :
e
1
+
h
=
e
1
+
u
=
e
×
e
u
=
e
×
(
1
+
u
+
u
2
2
+
u
3
6
+
o
(
u
3
)
)
=
e
×
(
1
+
(
h
2
−
h
2
8
+
h
3
16
)
+
1
2
(
(
h
2
)
2
−
2
h
2
h
2
8
)
+
1
6
(
h
2
)
3
+
o
(
h
3
)
)
=
e
×
(
1
+
h
2
+
h
3
48
)
+
o
(
h
3
)
{\displaystyle {\begin{aligned}\operatorname {e} ^{\sqrt {1+h}}&=\operatorname {e} ^{1+u}=\mathrm {e} \times \operatorname {e} ^{u}\\&=\mathrm {e} \times \left(1+u+{\frac {u^{2}}{2}}+{\frac {u^{3}}{6}}+o(u^{3})\right)\\&=\mathrm {e} \times \left(1+\left({\frac {h}{2}}-{\frac {h^{2}}{8}}+{\frac {h^{3}}{16}}\right)+{\frac {1}{2}}\left(\left({\frac {h}{2}}\right)^{2}-2{\frac {h}{2}}{\frac {h^{2}}{8}}\right)+{\frac {1}{6}}\left({\frac {h}{2}}\right)^{3}+o(h^{3})\right)\\&=\mathrm {e} \times \left(1+{\frac {h}{2}}+{\frac {h^{3}}{48}}\right)+o(h^{3})\end{aligned}}}
e
x
=
e
+
e
2
(
x
−
1
)
+
e
48
(
x
−
1
)
3
+
o
(
(
x
−
1
)
3
)
{\displaystyle \operatorname {e} ^{\sqrt {x}}=\mathrm {e} +{\frac {\mathrm {e} }{2}}(x-1)+{\frac {\mathrm {e} }{48}}(x-1)^{3}+o((x-1)^{3})}
On donne l'équation de van der Waals
(
P
+
a
V
2
)
(
V
−
b
)
=
T
{\displaystyle \left(P+{\frac {a}{V^{2}}}\right)\left(V-b\right)=T}
où
P
,
T
,
V
{\displaystyle P,T,V}
R
,
a
,
b
{\displaystyle R,a,b}
Quand
V
{\displaystyle V}
T
{\displaystyle T}
P
{\displaystyle P}
1
V
{\displaystyle {\frac {1}{V}}}
Solution
P
=
R
T
V
−
b
−
a
V
2
=
T
V
1
1
−
b
V
−
a
V
2
=
R
T
V
(
1
+
b
V
+
o
(
1
V
)
)
−
a
V
2
=
R
T
V
+
R
T
b
−
a
V
2
+
o
(
1
V
2
)
{\displaystyle P={\frac {RT}{V-b}}-{\frac {a}{V^{2}}}={\frac {T}{V}}{\frac {1}{1-{\frac {b}{V}}}}-{\frac {a}{V^{2}}}={\frac {RT}{V}}\left(1+{\frac {b}{V}}+o\left({\frac {1}{V}}\right)\right)-{\frac {a}{V^{2}}}={\frac {RT}{V}}+{\frac {RTb-a}{V^{2}}}+o\left({\frac {1}{V^{2}}}\right)}
Déterminer les asymptotes et les positions par rapport à ces asymptotes des courbes suivantes :
y
=
ln
cosh
x
{\displaystyle y=\ln \cosh x}
y
=
1
+
x
4
+
x
6
1
+
x
2
{\displaystyle y={\frac {\sqrt {1+x^{4}+x^{6}}}{1+x^{2}}}}
Étudier les fonctions suivantes :
f
(
x
)
=
x
ln
|
2
+
1
x
|
{\displaystyle f(x)=x\ln \left|2+{1 \over x}\right|}
g
(
x
)
=
(
1
+
1
x
)
x
{\displaystyle g(x)=(1+{1 \over x})^{x}}
h
(
x
)
=
(
x
−
1
x
)
e
1
/
x
{\displaystyle h(x)=(x-{1 \over x})\operatorname {e} ^{1/x}}
Solution
f
{\displaystyle f}
R
∖
{
0
,
−
1
2
}
{\displaystyle \mathbb {R} \setminus \{0,{-1 \over 2}\}}
f
(
x
)
=
x
(
ln
|
2
x
+
1
|
−
ln
|
x
|
)
∼
0
−
x
ln
|
x
|
→
0
{\displaystyle f(x)=x(\ln |2x+1|-\ln |x|)\sim _{0}-x\ln |x|\to 0}
x
{\displaystyle x}
x
→
0
{\displaystyle x\to 0}
(
0
,
0
)
{\displaystyle (0,0)}
lim
x
→
0
(
−
ln
|
x
|
)
=
+
∞
{\displaystyle \lim _{x\to 0}(-\ln |x|)=+\infty }
−
1
2
{\displaystyle {-1 \over 2}}
lim
−
1
2
f
=
−
1
2
ln
(
0
+
)
=
+
∞
{\displaystyle \lim _{-1 \over 2}f={-1 \over 2}\ln(0^{+})=+\infty }
x
=
−
1
2
{\displaystyle x={-1 \over 2}}
∞
{\displaystyle \infty }
f
(
x
)
=
x
[
ln
2
+
ln
(
1
+
1
2
x
)
]
=
x
[
ln
2
+
1
2
x
−
1
8
x
2
+
o
(
1
x
2
)
]
=
x
ln
2
+
1
2
−
1
8
x
+
o
(
1
x
)
{\displaystyle f(x)=x[\ln 2+\ln(1+{1 \over 2x})]=x[\ln 2+{1 \over 2x}-{1 \over 8x^{2}}+o({1 \over x^{2}})]=x\ln 2+{1 \over 2}-{1 \over 8x}+o({1 \over x})}
y
=
x
ln
2
+
1
2
{\displaystyle y=x\ln 2+{\frac {1}{2}}}
f
{\displaystyle f}
x
→
+
∞
{\displaystyle x\to +\infty }
x
→
−
∞
{\displaystyle x\to -\infty }
f
′
(
x
)
=
ln
|
2
+
1
x
|
−
1
2
x
+
1
{\displaystyle f'(x)=\ln \left|2+{1 \over x}\right|-{1 \over 2x+1}}
f
″
(
x
)
=
−
1
x
(
2
x
+
1
)
2
{\displaystyle f''(x)={-1 \over x(2x+1)^{2}}}
−
x
{\displaystyle -x}
sur
]
−
∞
,
−
1
2
[
{\displaystyle \left]-\infty ,{-1 \over 2}\right[}
f
′
{\displaystyle f'}
−
∞
{\displaystyle -\infty }
ln
2
>
0
{\displaystyle \ln 2>0}
f
′
>
0
{\displaystyle f'>0}
f
{\displaystyle f}
−
∞
{\displaystyle -\infty }
+
∞
{\displaystyle +\infty }
f
(
−
1
)
=
0
{\displaystyle f(-1)=0}
f
′
(
−
1
)
=
1
{\displaystyle f'(-1)=1}
sur
]
−
1
2
,
0
[
{\displaystyle \left]{-1 \over 2},0\right[}
f
′
{\displaystyle f'}
−
∞
{\displaystyle -\infty }
+
∞
{\displaystyle +\infty }
α
∈
]
−
1
2
,
0
[
{\displaystyle \alpha \in \left]{-1 \over 2},0\right[}
]
−
1
2
,
α
[
{\displaystyle \left]{-1 \over 2},\alpha \right[}
f
′
<
0
{\displaystyle f'<0}
f
{\displaystyle f}
+
∞
{\displaystyle +\infty }
f
(
α
)
{\displaystyle f(\alpha )}
f
(
−
1
3
)
=
0
{\displaystyle f({-1 \over 3})=0}
f
′
(
−
1
3
)
=
−
3
{\displaystyle f'\left({-1 \over 3}\right)=-3}
]
α
,
0
[
{\displaystyle \left]\alpha ,0\right[}
f
′
>
0
{\displaystyle f'>0}
f
{\displaystyle f}
f
(
α
)
{\displaystyle f(\alpha )}
0
{\displaystyle 0}
sur
]
0
,
+
∞
[
{\displaystyle \left]0,+\infty \right[}
f
′
{\displaystyle f'}
+
∞
{\displaystyle +\infty }
ln
2
>
0
{\displaystyle \ln 2>0}
f
′
>
0
{\displaystyle f'>0}
f
{\displaystyle f}
0
{\displaystyle 0}
+
∞
{\displaystyle +\infty }
g
(
x
)
=
e
k
(
x
)
{\displaystyle g(x)=\operatorname {e} ^{k(x)}}
k
(
x
)
=
x
ln
(
1
+
1
x
)
{\displaystyle k(x)=x\ln(1+{1 \over x})}
]
−
∞
,
−
1
[
∪
]
0
,
+
∞
[
{\displaystyle \left]-\infty ,-1\right[\cup \left]0,+\infty \right[}
0
+
{\displaystyle 0^{+}}
x
→
0
+
{\displaystyle x\to 0^{+}}
k
(
x
)
=
x
(
ln
(
1
+
x
)
−
ln
x
)
∼
0
+
−
x
ln
x
→
0
+
{\displaystyle k(x)=x(\ln(1+x)-\ln x)\sim _{0^{+}}-x\ln x\to 0^{+}}
g
(
x
)
→
1
+
{\displaystyle g(x)\to 1^{+}}
lim
(
−
1
)
−
k
=
−
1.
ln
(
0
+
)
=
+
∞
{\displaystyle \lim _{(-1)^{-}}k=-1.\ln(0^{+})=+\infty }
lim
(
−
1
)
−
g
=
+
∞
{\displaystyle \lim _{(-1)^{-}}g=+\infty }
k
(
x
)
∼
±
∞
x
(
1
x
)
=
1
{\displaystyle k(x)\sim _{\pm \infty }x({1 \over x})=1}
lim
±
∞
g
=
e
{\displaystyle \lim _{\pm \infty }g=\mathrm {e} }
g
′
=
g
×
k
′
{\displaystyle g'=g\times k'}
k
′
(
x
)
=
ln
(
1
+
1
x
)
−
1
1
+
x
>
0
{\displaystyle k'(x)=\ln \left(1+{1 \over x}\right)-{1 \over 1+x}>0}
k
″
(
x
)
=
−
1
(
1
+
x
)
2
.
x
{\displaystyle k''(x)={-1 \over (1+x)^{2}.x}}
−
x
{\displaystyle -x}
]
−
∞
,
−
1
[
{\displaystyle \left]-\infty ,-1\right[}
k
′
{\displaystyle k'}
−
∞
{\displaystyle -\infty }
0
{\displaystyle 0}
]
0
,
+
∞
[
{\displaystyle \left]0,+\infty \right[}
k
′
{\displaystyle k'}
+
∞
{\displaystyle +\infty }
0
{\displaystyle 0}
k
′
>
0
{\displaystyle k'>0}
g
{\displaystyle g}
0
+
{\displaystyle 0^{+}}
k
′
→
+
∞
{\displaystyle k'\to +\infty }
g
→
1
{\displaystyle g\to 1}
g
′
→
+
∞
{\displaystyle g'\to +\infty }
h
{\displaystyle h}
R
∗
{\displaystyle \mathbb {R} ^{*}}
h
(
x
)
=
(
x
−
1
x
)
(
1
+
1
x
+
1
2
x
2
+
o
(
1
x
2
)
)
=
x
+
1
−
1
2
x
+
o
(
1
x
)
{\displaystyle h(x)=(x-{1 \over x})(1+{1 \over x}+{1 \over 2x^{2}}+o({1 \over x^{2}}))=x+1-{1 \over 2x}+o({1 \over x})}
y
=
x
+
1
{\displaystyle y=x+1}
h
{\displaystyle h}
x
→
+
∞
{\displaystyle x\to +\infty }
x
→
−
∞
{\displaystyle x\to -\infty }
h
(
x
)
∼
0
−
1
x
e
1
x
{\displaystyle h(x)\sim _{0}{-1 \over x}\operatorname {e} ^{1 \over x}}
lim
0
+
h
=
−
lim
+
∞
u
e
u
=
−
∞
{\displaystyle \lim _{0^{+}}h=-\lim _{+\infty }u\operatorname {e} ^{u}=-\infty }
lim
0
−
h
=
lim
+
∞
u
e
−
u
=
0
+
{\displaystyle \lim _{0^{-}}h=\lim _{+\infty }u\operatorname {e} ^{-u}=0^{+}}
(
0
,
0
)
{\displaystyle (0,0)}
lim
x
→
0
−
h
(
x
)
x
=
lim
+
∞
−
e
−
u
=
0
{\displaystyle \lim _{x\to 0^{-}}{h(x) \over x}=\lim _{+\infty }-\operatorname {e} ^{-u}=0}
h
′
(
x
)
=
e
1
x
x
3
−
x
2
+
x
+
1
x
3
{\displaystyle h'(x)=\operatorname {e} ^{1 \over x}{x^{3}-x^{2}+x+1 \over x^{3}}}
x
3
−
x
2
+
x
+
1
{\displaystyle x^{3}-x^{2}+x+1}
−
∞
{\displaystyle -\infty }
+
∞
{\displaystyle +\infty }
>
0
{\displaystyle >0}
0
{\displaystyle 0}
<
0
{\displaystyle <0}
−
1
{\displaystyle -1}
α
∈
]
−
1
,
0
[
{\displaystyle \alpha \in \left]-1,0\right[}
α
∈
]
−
2
3
,
−
1
2
[
{\displaystyle \alpha \in \left]{-2 \over 3},{-1 \over 2}\right[}
h
{\displaystyle h}
]
−
∞
,
α
[
{\displaystyle \left]-\infty ,\alpha \right[}
]
0
,
+
∞
[
{\displaystyle \left]0,+\infty \right[}
]
α
,
0
[
{\displaystyle \left]\alpha ,0\right[}
h
{\displaystyle h}
±
1
{\displaystyle \pm 1}
h
′
(
−
1
)
=
2
e
{\displaystyle h'(-1)=2\mathrm {e} }
h
′
(
1
)
=
2
e
{\displaystyle h'(1)={2 \over \mathrm {e} }}
(
−
1
,
0
)
{\displaystyle (-1,0)}
h
″
(
x
)
=
−
e
1
x
x
2
+
4
x
+
1
x
5
{\displaystyle h''(x)=-\operatorname {e} ^{1 \over x}{x^{2}+4x+1 \over x^{5}}}
−
2
−
3
<
−
1
{\displaystyle -2-{\sqrt {3}}<-1}
−
2
+
3
∈
]
α
,
0
[
{\displaystyle -2+{\sqrt {3}}\in \left]\alpha ,0\right[}
On pose
f
(
x
)
=
x
3
1
+
x
6
{\displaystyle f(x)={\frac {x^{3}}{1+x^{6}}}}
f
{\displaystyle f}
f
(
n
)
(
0
)
{\displaystyle f^{(n)}(0)}
n
≥
0
{\displaystyle n\geq 0}
Soient
I
{\displaystyle I}
f
{\displaystyle f}
I
{\displaystyle I}
{
∀
x
∈
I
f
′
(
x
)
=
tan
(
x
+
f
(
x
)
)
f
(
0
)
=
π
/
4.
{\displaystyle {\begin{cases}\forall x\in I\quad f'(x)=\tan(x+f(x))\\f(0)=\pi /4.\end{cases}}}
Former le d.l. de
f
{\displaystyle f}
Solution
Par récurrence,
f
{\displaystyle f}
k k , donc C∞ .
f
′
=
tan
(
x
+
f
)
⇒
f
″
=
(
1
+
f
′
2
)
(
1
+
f
′
)
=
1
+
f
′
+
f
′
2
+
f
′
3
⇒
f
‴
=
f
″
(
1
+
2
f
′
+
3
f
′
2
)
⇒
f
(
4
)
=
f
‴
(
1
+
2
f
′
+
3
f
′
2
)
+
f
″
2
(
2
+
6
f
′
)
.
{\displaystyle {\begin{aligned}f'=\tan(x+f)&\Rightarrow f''=(1+f'^{2})(1+f')=1+f'+f'^{2}+f'^{3}\\&\Rightarrow f'''=f''(1+2f'+3f'^{2})\\&\Rightarrow f^{(4)}=f'''(1+2f'+3f'^{2})+f''^{2}(2+6f').\end{aligned}}}
On en déduit de proche en proche :
f
(
0
)
=
π
/
4
,
f
′
(
0
)
=
tan
(
0
+
π
/
4
)
=
1
,
f
″
(
0
)
=
(
1
+
1
2
)
(
1
+
1
)
=
4
,
f
‴
(
0
)
=
4
(
1
+
2.1
+
3.1
2
)
=
24
,
f
(
4
)
(
0
)
=
24
(
1
+
2.1
+
3.1
2
)
+
4
2
(
2
+
6.1
)
=
272
{\displaystyle f(0)=\pi /4,\;f'(0)=\tan(0+\pi /4)=1,\;f''(0)=(1+1^{2})(1+1)=4,\;f'''(0)=4(1+2.1+3.1^{2})=24,\;f^{(4)}(0)=24(1+2.1+3.1^{2})+4^{2}(2+6.1)=272}
d'où, par Taylor-Young :
f
(
x
)
=
π
4
+
x
+
4
2
!
x
2
+
24
3
!
x
3
+
272
4
!
x
4
+
o
(
x
4
)
=
π
4
+
x
+
2
x
2
+
4
x
3
+
34
3
x
4
+
o
(
x
4
)
{\displaystyle f(x)={\frac {\pi }{4}}+x+{\frac {4}{2!}}x^{2}+{\frac {24}{3!}}x^{3}+{\frac {272}{4!}}x^{4}+o(x^{4})={\frac {\pi }{4}}+x+2x^{2}+4x^{3}+{\frac {34}{3}}x^{4}+o(x^{4})}
« Calculateur en ligne de développements limités » , sur dcode.fr Pascal Lainé, « Développements limités, équivalents et calculs de limites » Arnaud Bodin, « Développements limités » , sur exo7 « Exercices corrigés - Formules de Taylor » , sur bibmath.net « Exercices corrigés - Développements limités » , sur bibmath.net « Math sup : Développements limités » , sur bibmath.net Patrice Lassère, « L2 Math » , septembre 2020
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